IODS course project
Annukka Puotiniemi
- 1. About the project
- 2. Regression and model validation
- 3. Logistic regression
- Part 1 - General Housekeeping
- Part 2 - The Data Set
- Part 3 - Choosing Variables of Interest
- Part 4 - Exploring the Variables
- Part 5 - Logistic Regression
- Part 6 - Exploring the Predictive Power of My Model
- Part 7 - Bonus: 10-fold Cross-validation
- Part 8 - Super-Bonus: Comparing the Performance of Different Logistic Regression Models
- 4. Clustering and Classification
- 5. Dimensionality reduction techniques
1. About the project
Introduction to Open Data Science
This 7-week course, as described in Kimmo Vehkalahti’s blog, is
open for everyone willing to learn how to use RStudio, R Markdown and GitHub (state-of-the-art tools of data science) to visualise and analyse open data with multivariate statistical methods following principles and practices of reproducible research
The course sounds exciting enough for me to cha(lle)nge my plans and become a data scientist!
2. Regression and model validation
Below are all the codes, my interpretations and explanations for this week’s data analysis exercises.
A brief note about data wrangling that preceded this diary entry. After I filtered my dataset to exclude rows that had zero values for points, almost the whole dataset printed NA values. I found the following solution in StackOverflow:
a_dataset <- analysis_dataset[apply(analysis_dataset!=0, 1, all),]
I now have the correct number of observations and variables, but this method would exclude all other zero rows too I think…
To read the previously wrangled data from my local folder into R:
mydata <- read.csv("learning2014.csv")The dataset is a combination of certain variables from a survey dataset. The survey examined the relationship between learning approaches and the achievements of students. All combination variables were scaled to the original scales. Observations where the exam points variable is zero (student did not sit the exam) were excluded to remove outliers. The (combination) variables included are:
- gender
- age
- attitude (Global attitude toward statistics)
- deep (measures deep learning)
- stra (measures strategic learning)
- surf (measures surface learning)
- points (points from the exam)
To begin the graphical overview of the data,I initialized the plot with data and aesthetic mappings, including colours by the variable gender, and showed summaries of the variables in the data:
library(ggplot2)
p1 <- ggplot(mydata, aes(x = attitude, y = points, col = gender))
p2 <- p1 + geom_point()
p3 <- p2 + geom_smooth(method = "lm")
p3
Summary of the data is presented below.
summary(mydata) ## X gender age attitude deep
## Min. : 1.00 F:110 Min. :17.00 Min. :1.400 Min. :1.583
## 1st Qu.: 44.25 M: 56 1st Qu.:21.00 1st Qu.:2.600 1st Qu.:3.333
## Median : 87.50 Median :22.00 Median :3.200 Median :3.667
## Mean : 90.13 Mean :25.51 Mean :3.143 Mean :3.680
## 3rd Qu.:136.75 3rd Qu.:27.00 3rd Qu.:3.700 3rd Qu.:4.083
## Max. :183.00 Max. :55.00 Max. :5.000 Max. :4.917
##
## stra surf points X.1
## Min. :1.250 Min. :1.583 Min. : 7.00 Min. :22.72
## 1st Qu.:2.625 1st Qu.:2.417 1st Qu.:19.00 1st Qu.:22.72
## Median :3.188 Median :2.833 Median :23.00 Median :22.72
## Mean :3.121 Mean :2.787 Mean :22.72 Mean :22.72
## 3rd Qu.:3.625 3rd Qu.:3.167 3rd Qu.:27.75 3rd Qu.:22.72
## Max. :5.000 Max. :4.333 Max. :33.00 Max. :22.72
## NA's :165
## X.2
## Min. :11.36
## 1st Qu.:11.36
## Median :11.36
## Mean :11.36
## 3rd Qu.:11.36
## Max. :11.36
## NA's :165The gender split of the respondents was 110 female and 56 male. Their ages ranged from 17 to 55, with an average age 25.51. The variable “attitude” is a combination variable of 10 variables and measures a global attitude toward statistics. The next three variables, deep approach (deep), strategic approach (stra) and surface approach (surf) are likewise combinatory variables and they have been scaled to the original scales. The average of attitude is 3.143, deep 3.68, strategic 3.121 and surface 2.787. The variable points shows the number of points the student achieved in an exam, and the average was 22.72.
For my model I chose the variables age, attitude and stra as explanatory variables and the target variable is points. A summary of the fitted model is shown below.
my_model <- lm(points ~ age + attitude + stra, data = mydata)
summary(my_model)##
## Call:
## lm(formula = points ~ age + attitude + stra, data = mydata)
##
## Residuals:
## Min 1Q Median 3Q Max
## -18.1149 -3.2003 0.3303 3.4129 10.7599
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 10.89543 2.64834 4.114 6.17e-05 ***
## age -0.08822 0.05302 -1.664 0.0981 .
## attitude 3.48077 0.56220 6.191 4.72e-09 ***
## stra 1.00371 0.53434 1.878 0.0621 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.26 on 162 degrees of freedom
## Multiple R-squared: 0.2182, Adjusted R-squared: 0.2037
## F-statistic: 15.07 on 3 and 162 DF, p-value: 1.07e-08The p-values for the coefficients indicate whether these relationships are statistically significant. The model above shows the variables with their significance levels. Age and strategic have a significance level of approximately 0.1 and 0.06 respectively, which, while being larger than the more or less generally accepted level of 0.05, still shows some significance. However, as keeping variables that are not statistically very significant can reduce the model’s precision, I will therefore only include attitude (with its significance code of 0) and stra (with the significance level that is very close to 0.05) in my re-fitted model. The new model is shown below.
my_model <- lm(points ~ attitude + stra, data = mydata)
summary(my_model)##
## Call:
## lm(formula = points ~ attitude + stra, data = mydata)
##
## Residuals:
## Min 1Q Median 3Q Max
## -17.6436 -3.3113 0.5575 3.7928 10.9295
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 8.9729 2.3959 3.745 0.00025 ***
## attitude 3.4658 0.5652 6.132 6.31e-09 ***
## stra 0.9137 0.5345 1.709 0.08927 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.289 on 163 degrees of freedom
## Multiple R-squared: 0.2048, Adjusted R-squared: 0.1951
## F-statistic: 20.99 on 2 and 163 DF, p-value: 7.734e-09The relationship between the target variable points and the explanatory variable attitude is a strong one as the model shows a significance code of 0. The relationship between points and strategic is also fairly strong. To explain how well my model fits the data, assessment of R-squared is required. R-squared indicates the percentage of the variance in the dependent variable points that the independent (explanatory) variables explain collectively. In this case the R-squared value is relatively low at approximately 20%, indicating that the fit is not very good. In fact, the multiple R-squared value was higher at approximately 22% in the model that I previously rejected.
The multiple regression model makes the following assumptions:
1. The relationship between the independent and dependent variables is linear.
2. The errors between observed and predicted values should be normally distributed.
3. The data has no multicollinearity (this happens when the independent variables correlate too highly with each other).
4. Homoscedasticity, a.k.a. homogeneity of variance.
The validity of these model assumptions can be explored by analyzing the residuals of the model, as in the plots shown below.
To produce the diagnostic plots Residuals vs Fitted values (plot 1), Normal QQ-plot (plot 2) and Residuals vs Leverage (plot 5):
par(mfrow = c(2,2))
plot(my_model, which = c(1:2, 5))
The normality assumption (assumption 2. above) can be explored by analyzing the Q-Q plot. The fit to the normality assumption is reasonably good in this case.
The homoscedasticity assumption (assumption 4.) implies that the size of the errors should not depend on the explanatory variables. This can be explored with the residuals vs fitted plot. As we can see from the first plot, there is a reasonably random placement of plots with no discernible pattern, and therefore the assumption is valid.
The last plot, residuals vs leverage, can help determine which observations have an unusually high impact on the model (assumption 3.). As we can observe from the plot, there is no single outlier that stands out, and therefore we can conclude that this assumption is valid.
After this graphical exploration we can conclude that the validity of the model assumptions is good.
3. Logistic regression
Moving on from linear regression, this week we will look at logictic regression. It is a method well suited for predicting and classifying data through probabilities. This week we will look at the concept of odds ratio (OR). We will also take a look at cross-validation, and learn about splitting the data into a training set and a test set.
Below are all the codes, my interpretations and explanations for this week’s data analysis exercises.
The data for this exercise was downloaded from the UC Irvine Machine Learning Repository.
The data are two student performance data sets that approach student achievement in secondary education of two schools in Portugal. The data variables include student grades, as well as demographic, social and school related features. The data was collated using school reports and questionnaires. The two datasets that analyze the performance in two distinct subjects: Mathematics (mat) and Portuguese language (por).
In the data wrangling exercise (which can be found in my GitHub repository) performed prior to this report, the two data sets were combined using a number of variables as student identifiers. Only the students present in both data sets were kept. Next, the average of the answers related to weekday and weekend alcohol consumption was taken to create a new variable ‘alc_use’. Then ‘alc_use’ was used to create a new logical column ‘high_use’ (TRUE for students whose ‘alc_use’ is greater than 2 and FALSE otherwise).
Citation:
P. Cortez and A. Silva. Using Data Mining to Predict Secondary School Student Performance. In A. Brito and J. Teixeira Eds., Proceedings of 5th FUture BUsiness TEChnology Conference (FUBUTEC 2008) pp. 5-12, Porto, Portugal, April, 2008, EUROSIS, ISBN 978-9077381-39-7.
Part 1 - General Housekeeping
For the first part of this analysis I created a new RMarkdown file and save it as an empty file named ‘chapter3.Rmd’. Then I included this file as a child file in my ‘index.Rmd’ file, and as a result you are now reading this.
Part 2 - The Data Set
I read the joined student alcohol consumption data into R from my local folder.
alc <- read.table("create_alc.csv", sep = "," , header=TRUE)Next, I printed out the names of the variables in the data.
colnames(alc)## [1] "school" "sex" "age" "address" "famsize"
## [6] "Pstatus" "Medu" "Fedu" "Mjob" "Fjob"
## [11] "reason" "nursery" "internet" "guardian" "traveltime"
## [16] "studytime" "failures" "schoolsup" "famsup" "paid"
## [21] "activities" "higher" "romantic" "famrel" "freetime"
## [26] "goout" "Dalc" "Walc" "health" "absences"
## [31] "G1" "G2" "G3" "alc_use" "high_use"The data set consists of 382 observations of 35 variables. The variables are a combination of numeric, nominal and binary attributes.
Part 3 - Choosing Variables of Interest
For this part, in order to study the relationship between high or low alcohol consumption, I chose the following 4 variables from the data, including my personal hypothesis for each variable:
1. age - I hypothesize that higher age will correlate with high alcohol consumption.
2. failures - indicating the number of past class failures. My hypothesis is that a high number of failures correlates with high alcohol use.
3. famrel - indicator for the quality of family relations. Better family relations will mean lower alcohol consumption and conversely worse relations will correlate with higher alcohol use.
4. absences - number of school absences. A higher number of absences will relate to high alcohol consumption.
Part 4 - Exploring the Variables
Next, I accessed some R packages that I will need shortly, and I also created a mini-dataset called alc_dataset. This mini-dataset will allow me to represent just the data I chose for my analysis.
Accessing packages dplyr and ggplot2:
library(dplyr); library(ggplot2)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionCreating alc_dataset:
alc_use <- alc$alc_use
high_use <- alc$high_use
age <- alc$age
failures <- alc$failures
famrel <- alc$famrel
absences <- alc$absences
alc_dataset <- data.frame(alc_use, age, failures, famrel, absences)
summary(alc_dataset)## alc_use age failures famrel
## Min. :1.000 Min. :15.00 Min. :0.0000 Min. :1.000
## 1st Qu.:1.000 1st Qu.:16.00 1st Qu.:0.0000 1st Qu.:4.000
## Median :1.500 Median :17.00 Median :0.0000 Median :4.000
## Mean :1.889 Mean :16.59 Mean :0.2016 Mean :3.937
## 3rd Qu.:2.500 3rd Qu.:17.00 3rd Qu.:0.0000 3rd Qu.:5.000
## Max. :5.000 Max. :22.00 Max. :3.0000 Max. :5.000
## absences
## Min. : 0.0
## 1st Qu.: 1.0
## Median : 3.0
## Mean : 4.5
## 3rd Qu.: 6.0
## Max. :45.0Above is the initial numeric summary for all the variables that I chose. This does not tell us anything when it comes to the relationships between the variables, but it is a nice neat summary of the distribution of each variable.
Let’s look at a simple plot that illustrates high_use.
ggplot(alc, aes(x = high_use, fill = sex)) +
geom_bar(position = "fill")
Unsurprisingly, most high users of alcohol in this dataset were male.
Before we get into the logistic regression in part 5 below, let’s look at some numerical (summaries by group using the pipe operator) and graphical (box plot) representations of the data. Here, I will compare the variable high_use with variables age, failures, famrel, and absences.
Variable age
alc %>% group_by(sex, high_use) %>% summarise(count = n(), mean_age = mean(age))## # A tibble: 4 x 4
## # Groups: sex [?]
## sex high_use count mean_age
## <fct> <lgl> <int> <dbl>
## 1 F FALSE 156 16.6
## 2 F TRUE 42 16.5
## 3 M FALSE 112 16.3
## 4 M TRUE 72 17.0# barplot
g1a <- ggplot(data=alc, aes(x=age, y=consumption)) +
geom_bar(stat="identity")
# initialise a plot of high_use and age
g1 <- ggplot(alc, aes(x = high_use, y = age, col = sex))
# define the plot as a boxplot and draw it
g1 + geom_boxplot() + ggtitle("Student ages by alcohol consumption and sex")
The numerical representation and the plot above quite clearly shows that my first hypothesis, concerning age, was largely rubbish. Moving on!
Variable failures
alc %>% group_by(sex, high_use) %>% summarise(count = n(), mean_failures = mean(failures))## # A tibble: 4 x 4
## # Groups: sex [?]
## sex high_use count mean_failures
## <fct> <lgl> <int> <dbl>
## 1 F FALSE 156 0.115
## 2 F TRUE 42 0.286
## 3 M FALSE 112 0.179
## 4 M TRUE 72 0.375# initialise a plot of high_use and failures
g2 <- ggplot(alc, aes(x = high_use, y = failures, col = sex))
# define the plot as a boxplot and draw it
g2 + geom_boxplot() + ggtitle("Student failures by alcohol consumption and sex")
The analysis of failures shows that, regardless of the rather useless box plot, my hypothesis was in the right track. In both males and females, the mean failures of those having high alcohol use was considerably more than the mean failures of those not having high alcohol use. In males, the difference between mean failures of high and low alcohol users, at 0.196, is higher than the difference, 0.171, in females. This is starting to point toward the fact that perhaps I should have hypothesized that being male is a factor in high alcohol use.
Variable famrel
alc %>% group_by(sex, high_use) %>% summarise(count = n(), mean_famrel = mean(famrel))## # A tibble: 4 x 4
## # Groups: sex [?]
## sex high_use count mean_famrel
## <fct> <lgl> <int> <dbl>
## 1 F FALSE 156 3.91
## 2 F TRUE 42 3.76
## 3 M FALSE 112 4.13
## 4 M TRUE 72 3.79# initialise a plot of high_use and family relationships
g3 <- ggplot(alc, aes(x = high_use, y = famrel, col = sex))
# define the plot as a boxplot and draw it
g3 + geom_boxplot() + ggtitle("Student family relationships by alcohol consumption and sex")
The analysis of famrel also shows that my hypothesis was not incorrect. In both sexes, the mean family relationships score of those having high alcohol use was lower than the mean failures of those not having high alcohol use. Here, a difference between the sexes is also evident, albeit in a manner rather different from the variable failures. The mean family relationship scores for the high alcohol users in both sexes was about the same, whereas, rather interrestingly, the family relationships score was better for males in the low alcohol use group.
Variable absences
alc %>% group_by(sex, high_use) %>% summarise(count = n(), mean_absences = mean(absences))## # A tibble: 4 x 4
## # Groups: sex [?]
## sex high_use count mean_absences
## <fct> <lgl> <int> <dbl>
## 1 F FALSE 156 4.22
## 2 F TRUE 42 6.79
## 3 M FALSE 112 2.98
## 4 M TRUE 72 6.12# initialise a plot of high_use and absences
g4 <- ggplot(alc, aes(x = high_use, y = absences, col = sex))
# define the plot as a boxplot and draw it
g4 + geom_boxplot() + ggtitle("Student absences by alcohol consumption and sex")
The analysis of absences shows that my hypothesis was correct: a higher number of absences does indeed correlate with high alcohol consumption. Again, the differences between the sexes in striking. The mean absences for both male and female high alcohol users are around the same ballpark, but the mean absence figures for low alcohol users differs dramatically between the sexes: 2.982 for the males and 4.224 for the females. This not only indicates that females generally have more school absences, but also that in males the introduction of high alcohol use dramatically increases the mean number of absences.
Part 5 - Logistic Regression
Next, I will use logistic regression to explore the relationship between my chosen variables and the binary variable high_use. This is done by using glm() to fit a logictic regression model. I will also print a summary of the model and I will use coef() to print out the coefficients of my model.
model <- glm(high_use ~ age + failures + famrel + absences, data = alc, family = "binomial")
summary(model)##
## Call:
## glm(formula = high_use ~ age + failures + famrel + absences,
## family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.2478 -0.7937 -0.6783 1.1488 1.9908
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -2.73938 1.73621 -1.578 0.114614
## age 0.13990 0.10248 1.365 0.172213
## failures 0.43635 0.18747 2.328 0.019933 *
## famrel -0.23849 0.12558 -1.899 0.057560 .
## absences 0.08040 0.02259 3.560 0.000371 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 434.98 on 377 degrees of freedom
## AIC: 444.98
##
## Number of Fisher Scoring iterations: 4coef(model)## (Intercept) age failures famrel absences
## -2.73937782 0.13989709 0.43634997 -0.23848933 0.08040375In the fitted model, absences shows a significance level ***, or less than 0.0001 (0.01%) and it is therefore statistically significant. failures is also significant with * a significance level of around 2%. famrel is close to the generally accepted significance level of 5%, whereas, as previously observed, age is not statistically significant.
As I observed above, it seems that sex has a statistical significance, and I am fitting my model again. This time I will include sex and exclude age.
model <- glm(high_use ~ sex + failures + famrel + absences, data = alc, family = "binomial")
summary(model)##
## Call:
## glm(formula = high_use ~ sex + failures + famrel + absences,
## family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.1174 -0.8376 -0.5867 1.0091 2.1557
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.83877 0.53446 -1.569 0.1166
## sexM 0.99120 0.24540 4.039 5.37e-05 ***
## failures 0.42458 0.18854 2.252 0.0243 *
## famrel -0.27632 0.12837 -2.153 0.0314 *
## absences 0.09052 0.02252 4.020 5.83e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 419.78 on 377 degrees of freedom
## AIC: 429.78
##
## Number of Fisher Scoring iterations: 4coef(model)## (Intercept) sexM failures famrel absences
## -0.83876936 0.99119526 0.42458409 -0.27631940 0.09051704As I suspected, the model now looks much better. All the variables have an acceptable level of significance, with the variable sexM (male gender) having a significance level on par with absences.
Now I will create the object OR (odds ratios) by using coef() on the model object to extract the coefficients of the model and then apply the exp function on the coefficients. The next step is to use confint() on the model to compute confidence intervals for the coefficients. I will exponentiate the values and assign the results to the object CI. Finally, I will combine cbind and print out the odds ratios and their confidence intervals.
OR <- coef(model) %>% exp
CI <- confint(model) %>% exp## Waiting for profiling to be done...cbind(OR, CI)## OR 2.5 % 97.5 %
## (Intercept) 0.4322421 0.1489271 1.2204360
## sexM 2.6944531 1.6760023 4.3948359
## failures 1.5289544 1.0573611 2.2280778
## famrel 0.7585706 0.5887336 0.9756953
## absences 1.0947402 1.0496217 1.1468498To be able to meaningfully interpret the coefficients of the model as odds ratios we need to know when odds ratios are used. They are used to compare the relative odds of the occurrence of the outcome of interest (in this case high alcohol use), given exposure to the variable of interest.
The odds ratios are interpreted as follows:
OR=1 Exposure does not affect odds of outcome OR>1 Exposure associated with higher odds of outcome OR<1 Exposure associated with lower odds of outcome
We therefore can see that exposure to famrel is associated with lower odds of outcome, whereas exposure to absences, failures, and especially sexM is associated with higher odds of the outcome high_use. sexM also has the widest confidence interval of all the variables.
If the confidence interval crosses 1 (contains the value of no effect), this implies that the observed effect is statistically not significant. This, as can be seen from the table above, is not the case for any of the variables.
As stated before, my hypothesis regarding the variable age was wrong, and as soon as it became apparent that a person being male was significant, I added the variable sexM to my model. The four variables in the fitted model all have a statistically significant relationship with high alcohol use in the following descending order: sexM, failures, absences, and famrel.
Part 6 - Exploring the Predictive Power of My Model
According to my logistic regression model, sexM, failures, famrel and absences all had a statistical relationship with high/low alcohol use. Therefore, I do not need to make any adjustments to my model at this stage - as I did so in the step above.
Below I explore the predictive power of my model by providing a 2x2 cross tabulation of predictions versus the actual values.
probabilities <- predict(model, type = "response")
alc <- mutate(alc, probability = probabilities)
alc <- mutate(alc, prediction = probability > 0.5)
select(alc, failures, absences, famrel, sex, high_use, probability, prediction) %>% tail(10)## failures absences famrel sex high_use probability prediction
## 373 1 0 4 M FALSE 0.3709210 FALSE
## 374 1 7 5 M TRUE 0.4573619 FALSE
## 375 0 1 5 F FALSE 0.1062293 FALSE
## 376 0 6 4 F FALSE 0.1976662 FALSE
## 377 1 2 5 F FALSE 0.1659304 FALSE
## 378 0 2 4 F FALSE 0.1464134 FALSE
## 379 2 2 2 F FALSE 0.4106677 FALSE
## 380 0 3 1 F FALSE 0.3007902 FALSE
## 381 0 4 2 M TRUE 0.4904650 FALSE
## 382 0 2 4 M TRUE 0.3160860 FALSEtable(high_use = alc$high_use, prediction = alc$prediction)## prediction
## high_use FALSE TRUE
## FALSE 253 15
## TRUE 82 32Let’s also draw a plot to examine the predictions and create a cross table of high_use versus prediction.
g <- ggplot(alc, aes(x = probability, y = high_use, col = prediction))
g + geom_point()
table(high_use = alc$high_use, prediction = alc$prediction) %>% prop.table %>% addmargins## prediction
## high_use FALSE TRUE Sum
## FALSE 0.66230366 0.03926702 0.70157068
## TRUE 0.21465969 0.08376963 0.29842932
## Sum 0.87696335 0.12303665 1.00000000The table above illustrates how accurate my predictions are. By adding the figures for high_use being FALSE while the model is predicting TRUE and high_use being TRUE while the model predicts FALSE we can calculate the percentage of all cases where my model would be wrong: 19.9%. Therefore, the model would be right roughly 80% of the time, which is much better a result than we would achieve by simple guessing with equal weights. The greatest delta between the number of observed and expected respondents is FALSE and FALSE at 63.9%, and my model comfortably outperforms this too, indicating that my model has some validity.
The next task at hand is accuracy and loss functions. I will compute the total proportion of inaccurately classified individuals, in other words the training error. The loss function loss_func is defined and called to compute the average number of wrong predictions in the (training) data.
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
loss_func(class = alc$high_use, prob = alc$probability)## [1] 0.2539267The training error for this model is approximately 0.254, while the test error is calculated in the following section.
Part 7 - Bonus: 10-fold Cross-validation
As a bonus I will perform a 10-old cross-validation on my model. This tells us the testing errors, as opposed to the training errors above.
# K-fold cross-validation
library(boot)
cv <- cv.glm(data = alc, cost = loss_func, glmfit = model, K = 10)
# average number of wrong predictions in the cross validation
cv$delta[1]## [1] 0.2617801My model has a very slightly better test set performance than the model that was introduced in DataCamp: 0.2617801 versus 0.2643979. The testing error of my model is higher than my training error. A lower training error is expected when a method easily overfits to the training data.
Part 8 - Super-Bonus: Comparing the Performance of Different Logistic Regression Models
As a super-bonus I will perform cross-validation to compare the performance of different logistic regression models. I will start with the following 20 predictors: sex, address, famsize, Pstatus, Medu, Fedu, Mjob, Fjob, guardian, studytime, failures, schoolsup, famsup, activities, higher, romantic, famrel, goout, health, and absences.
alc_use <- alc$alc_use
high_use <- alc$high_use
sex <- alc$sex
address <- alc$address
famsize <- alc$famsize
Pstatus <- alc$Pstatus
Medu <- alc$Medu
Fedu <- alc$Fedu
Mjob <- alc$Mjob
Fjob <- alc$Fjob
guardian <- alc$guardian
studytime <- alc$studytime
failures <- alc$failures
schoolsup <- alc$schoolsup
famsup <- alc$famsup
activities <- alc$activities
higher <- alc$higher
romantic <- alc$romantic
famrel <- alc$famrel
goout <- alc$goout
health <- alc$health
absences <- alc$absences
model_20 <- glm(high_use ~ sex + address + famsize + Pstatus + Medu + Fedu + Mjob + Fjob + guardian + studytime + failures + schoolsup + famsup + activities + higher + romantic + famrel + goout + health + absences, data = alc, family = "binomial")
probabilities <- predict(model_20, type = "response")
alc <- mutate(alc, probability = probabilities)
alc <- mutate(alc, prediction = probability > 0.5)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
t_er_20 <- loss_func(class = alc$high_use, prob = alc$probability)
t_er_20## [1] 0.1989529cv_20 <- cv.glm(data = alc, cost = loss_func, glmfit = model_20, K = 10)
ts_er_20 <- cv_20$delta[1]
ts_er_20## [1] 0.2434555Let’s see what happens when we reduce the number of predictors to 15.
I will leave out higher, famsup, schoolsup, Medu and Fedu from the next model, leaving me with 15 predictors.
model_15 <- glm(high_use ~ sex + address + famsize + Pstatus + Mjob + Fjob + guardian + studytime + failures + activities + romantic + famrel + goout + health + absences, data = alc, family = "binomial")
probabilities <- predict(model_15, type = "response")
alc <- mutate(alc, probability = probabilities)
alc <- mutate(alc, prediction = probability > 0.5)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
t_er_15 <- loss_func(class = alc$high_use, prob = alc$probability)
t_er_15## [1] 0.1937173cv_15 <- cv.glm(data = alc, cost = loss_func, glmfit = model_15, K = 10)
ts_er_15 <- cv_15$delta[1]
ts_er_15## [1] 0.2198953Now I will reduce the number of predictors to 10.
I will leave out famsize, Pstatus, Mjob, Fjob and romantic from the next model, leaving me with 10 predictors.
model_10 <- glm(high_use ~ sex + address + guardian + studytime + failures + activities + famrel + goout + health + absences, data = alc, family = "binomial")
probabilities <- predict(model_10, type = "response")
alc <- mutate(alc, probability = probabilities)
alc <- mutate(alc, prediction = probability > 0.5)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
t_er_10 <- loss_func(class = alc$high_use, prob = alc$probability)
t_er_10## [1] 0.2041885cv_10 <- cv.glm(data = alc, cost = loss_func, glmfit = model_10, K = 10)
ts_er_10 <- cv_10$delta[1]
ts_er_10## [1] 0.2251309Now I will reduce the number of predictors to 6.
I will take out 4 predictors (activities, guardian, failures and health).
model_6 <- glm(high_use ~ sex + address + studytime + famrel + goout + absences, data = alc, family = "binomial")
probabilities <- predict(model_6, type = "response")
alc <- mutate(alc, probability = probabilities)
alc <- mutate(alc, prediction = probability > 0.5)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
t_er_6 <- loss_func(class = alc$high_use, prob = alc$probability)
t_er_6## [1] 0.2172775cv_6 <- cv.glm(data = alc, cost = loss_func, glmfit = model_6, K = 10)
ts_er_6 <- cv_6$delta[1]
ts_er_6## [1] 0.2146597I will now take out the predictors one by one, until I reach 1.
model_5 <- glm(high_use ~ sex + studytime + famrel + goout + absences, data = alc, family = "binomial")
probabilities <- predict(model_5, type = "response")
alc <- mutate(alc, probability = probabilities)
alc <- mutate(alc, prediction = probability > 0.5)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
t_er_5 <- loss_func(class = alc$high_use, prob = alc$probability)
t_er_5## [1] 0.2172775cv_5 <- cv.glm(data = alc, cost = loss_func, glmfit = model_5, K = 10)
ts_er_5 <- cv_5$delta[1]
ts_er_5## [1] 0.2225131model_4 <- glm(high_use ~ sex + famrel + goout + absences, data = alc, family = "binomial")
probabilities <- predict(model_4, type = "response")
alc <- mutate(alc, probability = probabilities)
alc <- mutate(alc, prediction = probability > 0.5)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
t_er_4 <- loss_func(class = alc$high_use, prob = alc$probability)
t_er_4## [1] 0.2041885cv_4 <- cv.glm(data = alc, cost = loss_func, glmfit = model_4, K = 10)
ts_er_4 <- cv_4$delta[1]
ts_er_4## [1] 0.2120419model_3 <- glm(high_use ~ sex + goout + absences, data = alc, family = "binomial")
probabilities <- predict(model_3, type = "response")
alc <- mutate(alc, probability = probabilities)
alc <- mutate(alc, prediction = probability > 0.5)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
t_er_3 <- loss_func(class = alc$high_use, prob = alc$probability)
t_er_3## [1] 0.2094241cv_3 <- cv.glm(data = alc, cost = loss_func, glmfit = model_3, K = 10)
ts_er_3 <- cv_3$delta[1]
ts_er_3## [1] 0.2172775model_2 <- glm(high_use ~ sex + absences, data = alc, family = "binomial")
probabilities <- predict(model_2, type = "response")
alc <- mutate(alc, probability = probabilities)
alc <- mutate(alc, prediction = probability > 0.5)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
t_er_2 <- loss_func(class = alc$high_use, prob = alc$probability)
t_er_2## [1] 0.2565445cv_2 <- cv.glm(data = alc, cost = loss_func, glmfit = model_2, K = 10)
ts_er_2 <- cv_2$delta[1]
ts_er_2## [1] 0.2643979model_1 <- glm(high_use ~ sex, data = alc, family = "binomial")
probabilities <- predict(model_1, type = "response")
alc <- mutate(alc, probability = probabilities)
alc <- mutate(alc, prediction = probability > 0.5)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
t_er_1 <- loss_func(class = alc$high_use, prob = alc$probability)
t_er_1## [1] 0.2984293cv_1 <- cv.glm(data = alc, cost = loss_func, glmfit = model_1, K = 10)
ts_er_1 <- cv_1$delta[1]
ts_er_1## [1] 0.2984293Finally, I drew a simple graph that displays the trends of both training and testing errors by the number of predictors in the model.
t_er<-c(t_er_1,t_er_2,t_er_3,t_er_4,t_er_5,t_er_6,t_er_10,t_er_15,t_er_20)
ts_er<-c(ts_er_1,ts_er_2,ts_er_3,ts_er_4,ts_er_5,ts_er_6,ts_er_10,ts_er_15,ts_er_20)
plot(t_er, type="o", col="blue", xlab="Number of predictors / model complexity", ylab="Prediction error")
lines(ts_er, type="o", col="red")
axis(1, at=1:9, lab=c("1","2","3","4","5","6","10","15","20"))
title(main="Training and test errors by number of predictors")
legend("topright", c("training error","test error"),lty=c(1,1), lwd=c(2.5,2.5),col=c("blue","red")) 
Overfitting will occur as soon as the test error starts to increase while the training error is still decreasing. In my model the sweet spot is at 10 predictors, anything more than that and the model will overfit, possibly as it is too complex (possibly, because there are many more reasons for overfitting).
To the left of the graph are situations where the model is low in complexity and has high bias and low variance. To the right of the graph are situations where the model is high in complexity and has low bias and high variance.
That’s all for this week - I’m already looking forward to next week’s tasks!
4. Clustering and Classification
The topics of this week - clutering and classification - are visual tools for the exploration of statistical data. Clustering means that some data points are closer to each other than some other points: they are clustered. Once we have clustered successfully, we can try to classify new observations to these clusters, thus validating the results of clustering.
Below are all the codes, my interpretations and explanations for this week’s data analysis exercises.
Part 1 - General Housekeeping
As in last week, I created a new RMarkdown file and save it as an empty file named ‘chapter4.Rmd’. Then I included this file as a child file in my ‘index.Rmd’ file, and as a result you are now reading this.
I also accessed some packages that I might need later.
library(ggplot2); library(GGally); library(corrplot); library(tidyr); library(dplyr)##
## Attaching package: 'GGally'## The following object is masked from 'package:dplyr':
##
## nasa## corrplot 0.84 loadedPart 2 - Loading the Boston data
To load and explore the Boston data, I first needed to access the MASS package, and then load the “Boston” data. Below is the structure and summary of the data, and also a matrix of the variables.
# access the MASS package
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:dplyr':
##
## select# load the data
data("Boston")
# explore the dataset
str(Boston)## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...The Boston data set is a data frame with 506 observations of 14 variables. The data examines housing values in the suburbs of Boston. The variables are all numeric, with two variables being integer types.
A link to details about the Boston dataset can be seen here.
Here is the source of the data:
Harrison, D. and Rubinfeld, D.L. (1978) Hedonic prices and the demand for clean air. < em >J. Environ. Economics and Management < b >5, 81-102.
Belsley D.A., Kuh, E. and Welsch, R.E. (1980) < em >Regression Diagnostics. Identifying Influential Data and Sources of Collinearity. New York: Wiley.
Part 3 - Exploring the Boston data
Below is a graphical overview of the data and the summaries of the variables in the data.
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00pairs(Boston)
The variable of interest here, as the data looks at housing values, is medv. It shows the median value of owner-occupied homes in $1000s.
The values for medvvary from 5 to 50, witha mean value of 22.53. medv correlates quite well with all the variables (apart from the Charles River dummy variable), but most strongly with the average number of rooms per dwelling rm(0.695) and lower status of the population lstat(obviously with a negative correlation of -0.738).
Let’s look at a way of plotting the correlations, this time from the corrplotpackage. Also, the funstion cor() can be used to craete a correlation matrix of the data.
# calculate the correlation matrix and round it
cor_matrix<-cor(Boston) %>% round(digits = 2)
# print the correlation matrix
cor_matrix## crim zn indus chas nox rm age dis rad tax
## crim 1.00 -0.20 0.41 -0.06 0.42 -0.22 0.35 -0.38 0.63 0.58
## zn -0.20 1.00 -0.53 -0.04 -0.52 0.31 -0.57 0.66 -0.31 -0.31
## indus 0.41 -0.53 1.00 0.06 0.76 -0.39 0.64 -0.71 0.60 0.72
## chas -0.06 -0.04 0.06 1.00 0.09 0.09 0.09 -0.10 -0.01 -0.04
## nox 0.42 -0.52 0.76 0.09 1.00 -0.30 0.73 -0.77 0.61 0.67
## rm -0.22 0.31 -0.39 0.09 -0.30 1.00 -0.24 0.21 -0.21 -0.29
## age 0.35 -0.57 0.64 0.09 0.73 -0.24 1.00 -0.75 0.46 0.51
## dis -0.38 0.66 -0.71 -0.10 -0.77 0.21 -0.75 1.00 -0.49 -0.53
## rad 0.63 -0.31 0.60 -0.01 0.61 -0.21 0.46 -0.49 1.00 0.91
## tax 0.58 -0.31 0.72 -0.04 0.67 -0.29 0.51 -0.53 0.91 1.00
## ptratio 0.29 -0.39 0.38 -0.12 0.19 -0.36 0.26 -0.23 0.46 0.46
## black -0.39 0.18 -0.36 0.05 -0.38 0.13 -0.27 0.29 -0.44 -0.44
## lstat 0.46 -0.41 0.60 -0.05 0.59 -0.61 0.60 -0.50 0.49 0.54
## medv -0.39 0.36 -0.48 0.18 -0.43 0.70 -0.38 0.25 -0.38 -0.47
## ptratio black lstat medv
## crim 0.29 -0.39 0.46 -0.39
## zn -0.39 0.18 -0.41 0.36
## indus 0.38 -0.36 0.60 -0.48
## chas -0.12 0.05 -0.05 0.18
## nox 0.19 -0.38 0.59 -0.43
## rm -0.36 0.13 -0.61 0.70
## age 0.26 -0.27 0.60 -0.38
## dis -0.23 0.29 -0.50 0.25
## rad 0.46 -0.44 0.49 -0.38
## tax 0.46 -0.44 0.54 -0.47
## ptratio 1.00 -0.18 0.37 -0.51
## black -0.18 1.00 -0.37 0.33
## lstat 0.37 -0.37 1.00 -0.74
## medv -0.51 0.33 -0.74 1.00# visualize the correlation matrix
corrplot(cor_matrix, method="square", type="lower", cl.pos="b", tl.pos="d", tl.cex = 0.6)
The corrplotis a really neat visualization method. Positive correlations are displayed in blue and negative correlations in red color. Color intensity and the size of the square are proportional to the correlation coefficients. As we can see from above, this gives us a quick visual way of confirming what we knew from before lstathas a strong negative correlation with medvand rm has a strong positive correlation with medv. At a glance, we can see that there are a handful of other strong positive and negative correlations in the matrix too.
Part 4 - Creating Training and Test Sets
Now we need to scale the data. This is done by subtracting the column means from the corresponding columns and dividing the difference with standard deviation. \[scaled(x)=\frac{x-mean(x)}{sd(x)}\] The dataset is then standardized, and below I have printed out a summary of the scaled data.
# center and standardize variables
boston_scaled <- scale(Boston)
# summaries of the scaled variables
summary(boston_scaled)## crim zn indus
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668
## Median :-0.390280 Median :-0.48724 Median :-0.2109
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202
## chas nox rm age
## Min. :-0.2723 Min. :-1.4644 Min. :-3.8764 Min. :-2.3331
## 1st Qu.:-0.2723 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366
## Median :-0.2723 Median :-0.1441 Median :-0.1084 Median : 0.3171
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.:-0.2723 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059
## Max. : 3.6648 Max. : 2.7296 Max. : 3.5515 Max. : 1.1164
## dis rad tax ptratio
## Min. :-1.2658 Min. :-0.9819 Min. :-1.3127 Min. :-2.7047
## 1st Qu.:-0.8049 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876
## Median :-0.2790 Median :-0.5225 Median :-0.4642 Median : 0.2746
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6617 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058
## Max. : 3.9566 Max. : 1.6596 Max. : 1.7964 Max. : 1.6372
## black lstat medv
## Min. :-3.9033 Min. :-1.5296 Min. :-1.9063
## 1st Qu.: 0.2049 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median : 0.3808 Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4332 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 0.4406 Max. : 3.5453 Max. : 2.9865# class of the boston_scaled object
class(boston_scaled)## [1] "matrix"# change the object to data frame
boston_scaled <- as.data.frame(boston_scaled)How did the variables change? They have changed per the scaled(x) equation above. It is worth noting that in this new summary, all the mean values are 0.
We can create a categorical variable from a continuous one, and we will do that with the crime rate in the Boston dataset (from the scaled crime rate). We will cut the crimvariable by quantiles to get the high, low and middle rates of crime into their own categories. The quantiles are used as the break points in the new categorical variable. The old crime rate variable is dropped from the dataset.
# summary of the scaled crime rate
summary(boston_scaled$crim)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -0.419367 -0.410563 -0.390280 0.000000 0.007389 9.924110# create a quantile vector of crim and print it
bins <- quantile(boston_scaled$crim)
bins## 0% 25% 50% 75% 100%
## -0.419366929 -0.410563278 -0.390280295 0.007389247 9.924109610# create a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, labels = c("low", "med_low", "med_high", "high"))
# look at the table of the new factor crime
table(crime)## crime
## low med_low med_high high
## 127 126 126 127# remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)
# add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)Next, we will divide the dataset to training and test sets, so that 80% of the data belongs to the training set.
# number of rows in the Boston dataset
n <- nrow(boston_scaled)
# choose randomly 80% of the rows
ind <- sample(n, size = n * 0.8)
# create train set
train <- boston_scaled[ind,]
# create test set
test <- boston_scaled[-ind,]
# save the correct classes from test data
correct_classes <- test$crime
# remove the crime variable from test data
test <- dplyr::select(test, -crime)Part 5 - Linear Discriminant Analysis
Next we will look at linear discriminant analysis. It is a classification and dimension reduction method that is closely related to logictic regression (from last week) and principal component analysis (next week). It can be used to find variables that discriminate or separate the classes best, or it can be used to predict the classes of new data.
First, we will fit the linear discriminant analysis on the training set created in the previous step. We will use the categorical crime rate as the target variable and all the other variables are predictor variables. Then we will draw the LDA (bi)plot.
# linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)
# print the lda.fit object
lda.fit## Call:
## lda(crime ~ ., data = train)
##
## Prior probabilities of groups:
## low med_low med_high high
## 0.2524752 0.2549505 0.2475248 0.2450495
##
## Group means:
## zn indus chas nox rm
## low 1.0485600 -0.9043744 -0.07933396 -0.8834533 0.46505807
## med_low -0.1028029 -0.2554402 0.03346513 -0.5675294 -0.15038562
## med_high -0.3997708 0.2056048 0.23949396 0.3612856 0.06132161
## high -0.4872402 1.0171737 -0.07348562 1.0920772 -0.47363761
## age dis rad tax ptratio
## low -0.9219333 0.9080489 -0.6947544 -0.7120204 -0.45132878
## med_low -0.3909571 0.3555077 -0.5603950 -0.5077633 -0.09695511
## med_high 0.4059169 -0.3536801 -0.3881138 -0.2804555 -0.23119916
## high 0.7989934 -0.8521952 1.6375616 1.5136504 0.78011702
## black lstat medv
## low 0.37926223 -0.78042504 0.527290609
## med_low 0.34164288 -0.13634468 0.004878011
## med_high 0.07393481 -0.03503166 0.138651652
## high -0.86962409 0.90998374 -0.751167306
##
## Coefficients of linear discriminants:
## LD1 LD2 LD3
## zn 0.08978308 0.8779497555 -0.82763268
## indus -0.04492351 -0.2201202906 0.42109419
## chas -0.06447413 -0.0662179563 0.07140265
## nox 0.45386686 -0.5659560859 -1.41594546
## rm -0.13672902 -0.0001595234 -0.19009080
## age 0.29636691 -0.3106487726 -0.30206077
## dis -0.01581766 -0.3010768532 0.11309579
## rad 2.93442346 0.9086892282 0.37654788
## tax 0.02122966 -0.0979694540 0.15013775
## ptratio 0.14180216 0.0742824981 -0.33403111
## black -0.16609864 -0.0060968320 0.13348591
## lstat 0.24892361 -0.1745741765 0.39123456
## medv 0.22918340 -0.3749878305 -0.24767960
##
## Proportion of trace:
## LD1 LD2 LD3
## 0.9431 0.0416 0.0152Now, I will try to explain the above LDA model output with the help of a very useful video that can be found here. Look for the part starting at 02:46.
First at the top we have the prior probabilities of groups. These are simply the number of observations in each class divided by the number of observations in the whole dataset.
The group means have the value for every variable and for every class. The means differ between the classes.
Then there are the coefficients of the linear discriminants. There is a coefficient for each of the variables. We have four target variables, and therefore three linear discriminants.
The proportion of trace is the between-group variance. In our model, Linear Discriminant 1 explains almost 95% of the between-group variance.
# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "orange", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(train$crime)
# plot the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1.5)
Part 6 - Predict with the LDA Model
The data was split earlier so that we now have the test set and the correct class labels. Now we will predict the classes with the LDA model on the test data. Based on the trained model, LDA calculates the probabilites for the new observation for belonging in each of the classes. The observation is classified to the class of the highest probability. The probabilities are estimated using Bayes Theorem.
The results are cross tabulated with the crime categories from the test set.
Save the crime categories from the test set and then remove the categorical crime variable from the test dataset
# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)
# cross tabulate the results
tt <- table(correct = correct_classes, predicted = lda.pred$class)
tt## predicted
## correct low med_low med_high high
## low 9 16 0 0
## med_low 3 12 8 0
## med_high 1 4 21 0
## high 0 0 0 28error = sum(tt[row(tt) != col(tt)]) / sum(tt)
error## [1] 0.3137255summary(test)## zn indus chas
## Min. :-0.48724 Min. :-1.40179 Min. :-0.2723
## 1st Qu.:-0.48724 1st Qu.:-0.89526 1st Qu.:-0.2723
## Median :-0.48724 Median :-0.37560 Median :-0.2723
## Mean :-0.07991 Mean :-0.02651 Mean :-0.1179
## 3rd Qu.:-0.48724 3rd Qu.: 1.01499 3rd Qu.:-0.2723
## Max. : 3.37170 Max. : 2.42017 Max. : 3.6648
## nox rm age
## Min. :-1.38676 Min. :-2.35780 Min. :-2.1591
## 1st Qu.:-0.82151 1st Qu.:-0.54316 1st Qu.:-0.4849
## Median :-0.23037 Median :-0.10053 Median : 0.3295
## Mean : 0.04239 Mean : 0.08639 Mean : 0.1433
## 3rd Qu.: 0.64339 3rd Qu.: 0.60398 3rd Qu.: 0.9610
## Max. : 2.72965 Max. : 3.47325 Max. : 1.1164
## dis rad tax
## Min. :-1.24706 Min. :-0.98187 Min. :-1.31269
## 1st Qu.:-0.81659 1st Qu.:-0.63733 1st Qu.:-0.73715
## Median :-0.42439 Median :-0.52248 Median :-0.38115
## Mean :-0.09317 Mean : 0.05175 Mean : 0.03059
## 3rd Qu.: 0.48305 3rd Qu.: 1.65960 3rd Qu.: 1.52941
## Max. : 3.95660 Max. : 1.65960 Max. : 1.79642
## ptratio black lstat
## Min. :-2.70470 Min. :-3.82271 Min. :-1.36017
## 1st Qu.:-0.74160 1st Qu.: 0.22198 1st Qu.:-0.73911
## Median : 0.34387 Median : 0.39587 Median :-0.08865
## Mean : 0.01873 Mean : 0.04731 Mean : 0.06923
## 3rd Qu.: 0.80578 3rd Qu.: 0.42977 3rd Qu.: 0.75611
## Max. : 1.26768 Max. : 0.44062 Max. : 3.04114
## medv
## Min. :-1.54753
## 1st Qu.:-0.62333
## Median :-0.21559
## Mean : 0.06092
## 3rd Qu.: 0.26826
## Max. : 2.98650As the cross tabulation shows, the LDA model performs exceptionally well only for high when predicting on new (test) data. For rest of the classes, the predictions are roughly 50% - 60% correct. Overall, the error rate is 33%. It is worth noting here that because of the way that the observations are divided randomly, we will see a slightly different table every time we run the function.
Part 7 - K-means Algorithm
Now we will look at clustering, and we will start with distance measures. First, we need to reload the Boston dataset and standardize the dataset.
library(MASS)
data('Boston')Now we will scale the Bostonvariables to get comparable distances.
# center and standardize variables
boston_K_scaled <- scale(Boston)
# summaries of the scaled variables
summary(boston_K_scaled)## crim zn indus
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668
## Median :-0.390280 Median :-0.48724 Median :-0.2109
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202
## chas nox rm age
## Min. :-0.2723 Min. :-1.4644 Min. :-3.8764 Min. :-2.3331
## 1st Qu.:-0.2723 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366
## Median :-0.2723 Median :-0.1441 Median :-0.1084 Median : 0.3171
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.:-0.2723 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059
## Max. : 3.6648 Max. : 2.7296 Max. : 3.5515 Max. : 1.1164
## dis rad tax ptratio
## Min. :-1.2658 Min. :-0.9819 Min. :-1.3127 Min. :-2.7047
## 1st Qu.:-0.8049 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876
## Median :-0.2790 Median :-0.5225 Median :-0.4642 Median : 0.2746
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6617 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058
## Max. : 3.9566 Max. : 1.6596 Max. : 1.7964 Max. : 1.6372
## black lstat medv
## Min. :-3.9033 Min. :-1.5296 Min. :-1.9063
## 1st Qu.: 0.2049 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median : 0.3808 Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4332 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 0.4406 Max. : 3.5453 Max. : 2.9865# class of the boston_scaled object
class(boston_K_scaled)## [1] "matrix"# change the object to data frame
boston_K_scaled <- as.data.frame(boston_K_scaled)Now we will calculate the distances between the observations. We will use the dist()function for this, and by default it will use Euclidean distance measure to create a distance matrix. We will also look at the Manhattan method.
# euclidean distance matrix
dist_eu <- dist(boston_K_scaled)
# look at the summary of the distances
summary(dist_eu)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.1343 3.4625 4.8241 4.9111 6.1863 14.3970# manhattan distance matrix
dist_man <- dist(boston_K_scaled, method = 'manhattan')
# look at the summary of the distances
summary(dist_man)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.2662 8.4832 12.6090 13.5488 17.7568 48.8618K-means is one of the best known clustering methods, and it is an unsupervised one. It assigns obsrvations to clusters based on the similarities of the objects. We will next run a k-means algorithm on the dataset. We will use 4 centers to begin with. This could in theory be any number though, hence the K.
# k-means clustering
km <-kmeans(boston_K_scaled, centers = 4)
# plot the Boston dataset with clusters
pairs(boston_K_scaled, col = km$cluster)
But now the question arises: what is the optimal number of clusters? In other words, we need to determine the K. There are more than one way to skin the proverbial cat here, but we will look at how the total of within cluster sum of squares or WCSS behaves when we change the number of clusters. The WCSS is calculated as follows:
\(WCSS=\sum^N_{i} (X_i-centroid)^2\)
We will achieve this by plotting the number of clusters and the total WCSS. Once the WCSS drops radically, we have the optimal number of clusters. So, let’s see where we get our radical drop!
set.seed(123) #K-means might produce different results every time, because it randomly assigns the initial cluster centers. This function is used to deal with that.
# determine the number of clusters
k_max <- 10
# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(boston_K_scaled, k)$tot.withinss})
# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')
In this case, two clusters seems optimal. We will therefore run kmeans()again with two clusters and visualize the results. I will split the pairs plot into two separate ones, for more readable results. I was going to use ggpairs but drawing the plot took soooooo long!
# k-means clustering
km <-kmeans(boston_K_scaled, centers = 2, nstart = 20)
km## K-means clustering with 2 clusters of sizes 177, 329
##
## Cluster means:
## crim zn indus chas nox rm
## 1 0.7238295 -0.4872402 1.1425514 -0.005407018 1.0824279 -0.4547830
## 2 -0.3894158 0.2621323 -0.6146857 0.002908943 -0.5823397 0.2446705
## age dis rad tax ptratio black
## 1 0.8051309 -0.8439539 1.0834228 1.1693521 0.5471636 -0.6101842
## 2 -0.4331555 0.4540421 -0.5828749 -0.6291043 -0.2943707 0.3282754
## lstat medv
## 1 0.8421083 -0.6566834
## 2 -0.4530491 0.3532917
##
## Clustering vector:
## 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2
## 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 2
## 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144
## 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162
## 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 2 2
## 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
## 2 2 2 1 2 1 2 2 1 1 2 2 2 2 2 2 2 2
## 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1
## 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378
## 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396
## 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414
## 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432
## 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450
## 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468
## 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486
## 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## 487 488 489 490 491 492 493 494 495 496 497 498 499 500 501 502 503 504
## 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2
## 505 506
## 2 2
##
## Within cluster sum of squares by cluster:
## [1] 1890.637 2686.045
## (between_SS / total_SS = 35.3 %)
##
## Available components:
##
## [1] "cluster" "centers" "totss" "withinss"
## [5] "tot.withinss" "betweenss" "size" "iter"
## [9] "ifault"# plot the Boston dataset with clusters
pairs(boston_K_scaled[1:5], col = km$cluster)
pairs(boston_K_scaled[6:10], col = km$cluster)
The first cluster is composed of 177 observations, and the second of 329. Only 35.3% of the total variance in the data set is explained by the clustering, and this indicates a poor fit.
Bonus
Perform k-means on the original Boston data with some reasonable number of clusters (> 2) (I did 3). Remember to standardize the dataset.
library(MASS)
data('Boston')# center and standardize variables
boston_K2_scaled <- scale(Boston)
# summaries of the scaled variables
summary(boston_K2_scaled)## crim zn indus
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668
## Median :-0.390280 Median :-0.48724 Median :-0.2109
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202
## chas nox rm age
## Min. :-0.2723 Min. :-1.4644 Min. :-3.8764 Min. :-2.3331
## 1st Qu.:-0.2723 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366
## Median :-0.2723 Median :-0.1441 Median :-0.1084 Median : 0.3171
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.:-0.2723 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059
## Max. : 3.6648 Max. : 2.7296 Max. : 3.5515 Max. : 1.1164
## dis rad tax ptratio
## Min. :-1.2658 Min. :-0.9819 Min. :-1.3127 Min. :-2.7047
## 1st Qu.:-0.8049 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876
## Median :-0.2790 Median :-0.5225 Median :-0.4642 Median : 0.2746
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6617 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058
## Max. : 3.9566 Max. : 1.6596 Max. : 1.7964 Max. : 1.6372
## black lstat medv
## Min. :-3.9033 Min. :-1.5296 Min. :-1.9063
## 1st Qu.: 0.2049 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median : 0.3808 Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4332 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 0.4406 Max. : 3.5453 Max. : 2.9865# class of the boston_scaled object
class(boston_K2_scaled)## [1] "matrix"# change the object to data frame
boston_K2_scaled <- as.data.frame(boston_K2_scaled)
# kmeans with 3 clusters
km <-kmeans(boston_K2_scaled, centers = 3, nstart = 20)
km## K-means clustering with 3 clusters of sizes 164, 236, 106
##
## Cluster means:
## crim zn indus chas nox rm
## 1 0.8046456 -0.4872402 1.117990 0.01575144 1.1253988 -0.46443119
## 2 -0.3760908 -0.3417123 -0.296848 0.01127561 -0.3345884 -0.09228038
## 3 -0.4075892 1.5146367 -1.068814 -0.04947434 -0.9962503 0.92400834
## age dis rad tax ptratio black
## 1 0.79737580 -0.85425848 1.2219249 1.2954050 0.60580719 -0.6407268
## 2 -0.02966623 0.05695857 -0.5803944 -0.6030198 -0.08691245 0.2863040
## 3 -1.16762641 1.19486951 -0.5983266 -0.6616391 -0.74378342 0.3538816
## lstat medv
## 1 0.8719904 -0.68418954
## 2 -0.1801190 0.03577844
## 3 -0.9480974 0.97889973
##
## Clustering vector:
## 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
## 2 2 2 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2
## 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
## 2 2 2 3 3 3 2 2 2 2 2 2 2 2 2 2 3 3
## 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72
## 3 3 3 3 3 2 2 2 2 3 3 3 3 2 2 2 2 2
## 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90
## 2 2 2 2 2 2 2 2 3 2 3 2 2 2 2 2 2 2
## 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108
## 2 2 2 2 2 2 2 3 3 2 2 2 2 2 2 2 2 2
## 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144
## 2 1 1 1 2 2 2 1 1 1 1 1 1 1 1 1 1 1
## 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162
## 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2
## 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
## 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198
## 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3
## 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216
## 3 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2
## 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234
## 2 2 2 2 2 2 2 2 3 3 2 2 3 2 2 2 3 3
## 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252
## 2 2 2 2 3 3 3 2 3 3 2 2 3 2 3 3 3 3
## 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270
## 3 3 3 3 3 3 2 2 2 3 3 2 2 2 2 3 3 2
## 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288
## 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
## 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306
## 3 3 3 3 3 2 2 2 2 2 3 3 3 3 3 3 3 2
## 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324
## 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
## 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342
## 2 2 2 2 2 2 2 3 3 2 2 2 2 2 2 2 2 3
## 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360
## 2 3 3 2 2 3 3 3 3 3 3 3 3 3 1 1 1 1
## 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378
## 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396
## 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414
## 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432
## 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450
## 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468
## 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486
## 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## 487 488 489 490 491 492 493 494 495 496 497 498 499 500 501 502 503 504
## 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2
## 505 506
## 2 2
##
## Within cluster sum of squares by cluster:
## [1] 1717.6354 1394.9385 758.7999
## (between_SS / total_SS = 45.2 %)
##
## Available components:
##
## [1] "cluster" "centers" "totss" "withinss"
## [5] "tot.withinss" "betweenss" "size" "iter"
## [9] "ifault"Then perform LDA using the clusters as target classes. Include all the variables in the Boston data in the LDA model.
So clusters are my TARGET classes, therefore the clusters need to be the variables before ~. So I need to somehow combine km$centers(??) and the Boston data… I think!
# access the cluster component in the kmeans data and change it to dataframe
cluster <- km$cluster
cluster <- as.data.frame(cluster)
# Add cluster as a column to boston_K2_scaled, ie. merge the two datasets
bonus_data <- merge(boston_K2_scaled, cluster)
# Create training and test sets
# number of rows in the new dataset
n <- nrow(bonus_data)
# choose randomly 80% of the rows
ind <- sample(n, size = n * 0.8)
# create train set
bonus_train <- bonus_data[ind,]
# create test set
bonus_test <- bonus_data[-ind,]
# Perform LDA
bonus.lda.fit <- lda(cluster ~ ., data = bonus_train)
bonus.lda.fit## Call:
## lda(cluster ~ ., data = bonus_train)
##
## Prior probabilities of groups:
## 1 2 3
## 0.3240231 0.4665036 0.2094733
##
## Group means:
## crim zn indus chas nox
## 1 0.0016554458 -0.0032816859 0.003320093 0.0016763564 0.002492146
## 2 -0.0001328571 0.0001191164 -0.001702143 0.0033211011 -0.001328476
## 3 0.0027073814 0.0032566259 -0.002699963 -0.0003492724 -0.003361992
## rm age dis rad tax
## 1 -0.0007286982 0.0021938001 -0.0047324208 0.0031777545 0.001790359
## 2 0.0017871229 -0.0004809195 0.0006946673 -0.0008323246 -0.001622667
## 3 0.0004078742 0.0003715072 0.0015532537 -0.0010301611 -0.002173667
## ptratio black lstat medv
## 1 0.002048162 -0.0019099538 0.001157256 -9.782444e-04
## 2 -0.001814217 -0.0004205507 -0.001427496 2.688129e-03
## 3 0.001119350 0.0029583281 -0.000103205 8.492379e-05
##
## Coefficients of linear discriminants:
## LD1 LD2
## crim 0.55072337 0.01850667
## zn 0.44711852 0.76863579
## indus -0.38179330 0.47500311
## chas -0.08762222 -0.31334240
## nox -0.22621821 -0.43228909
## rm 0.02071274 0.01911782
## age 0.71168450 0.04897999
## dis 0.44531388 -1.10905794
## rad -0.61478741 1.02250037
## tax 0.32188018 -1.24967843
## ptratio 0.19325044 0.28918008
## black 0.36096685 0.36722420
## lstat 0.24613638 -0.41377554
## medv 0.10176888 -0.81631237
##
## Proportion of trace:
## LD1 LD2
## 0.5699 0.4301Visualize the results with a biplot (include arrows representing the relationships of the original variables to the LDA solution). Interpret the results. Which variables are the most influencial linear separators for the clusters?
# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "orange", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(bonus_train$cluster)
# plot the lda results
plot(bonus.lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(bonus.lda.fit, myscale = 1.5)
Super-Bonus
Run the code below for the (scaled) train data that you used to fit the LDA. The code creates a matrix product, which is a projection of the data points.
model_predictors <- dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)## [1] 404 13dim(lda.fit$scaling)## [1] 13 3# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)Next, install and access the plotly package. Create a 3D plot (Cool!) of the columns of the matrix product by typing the code below.
library(plotly)##
## Attaching package: 'plotly'## The following object is masked from 'package:MASS':
##
## select## The following object is masked from 'package:ggplot2':
##
## last_plot## The following object is masked from 'package:stats':
##
## filter## The following object is masked from 'package:graphics':
##
## layoutplot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers')No kidding, this is cool!!
Adjust the code: add argument color as a argument in the plot_ly() function. Set the color to be the crime classes of the train set. Draw another 3D plot where the color is defined by the clusters of the k-means. How do the plots differ? Are there any similarities? (0-3 points to compensate any loss of points from the above exercises)
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', col=bins)## Warning: 'scatter3d' objects don't have these attributes: 'col'
## Valid attributes include:
## 'type', 'visible', 'showlegend', 'legendgroup', 'opacity', 'name', 'uid', 'ids', 'customdata', 'selectedpoints', 'hoverinfo', 'hoverlabel', 'stream', 'transforms', 'x', 'y', 'z', 'text', 'hovertext', 'mode', 'surfaceaxis', 'surfacecolor', 'projection', 'connectgaps', 'line', 'marker', 'textposition', 'textfont', 'error_x', 'error_y', 'error_z', 'xcalendar', 'ycalendar', 'zcalendar', 'scene', 'idssrc', 'customdatasrc', 'hoverinfosrc', 'xsrc', 'ysrc', 'zsrc', 'textsrc', 'hovertextsrc', 'textpositionsrc', 'key', 'set', 'frame', 'transforms', '_isNestedKey', '_isSimpleKey', '_isGraticule', '_bbox'5. Dimensionality reduction techniques
Part 1. Overview of the Data
For clarity, here are the variables and their short explanations:
- Life.Exp = Life expectancy at birth - Edu.Exp = Expected years of schooling - GNI = Gross National Income - Mat.Mor = Maternal mortality ratio - Ado.Birth = Adolescent birth rate - Parli.F = Percetange of female representatives in parliament - Edu2.FM = Proportion of females with at least secondary education / Proportion of males with at least secondary education - Labo.FM = Proportion of females in the labour force / Proportion of males in the labour force
# dplyr, corrplot and GGally are available
library(dplyr)
library(corrplot)
library(GGally)
human <- read.table("create_human.csv", sep = "," , header=TRUE)
human <- select(human, -1)
summary(human)## Edu2.FM Labo.FM Life.Exp Edu.Exp
## Min. :0.1717 Min. :0.1857 Min. :49.00 Min. : 5.40
## 1st Qu.:0.7264 1st Qu.:0.5984 1st Qu.:66.30 1st Qu.:11.25
## Median :0.9375 Median :0.7535 Median :74.20 Median :13.50
## Mean :0.8529 Mean :0.7074 Mean :71.65 Mean :13.18
## 3rd Qu.:0.9968 3rd Qu.:0.8535 3rd Qu.:77.25 3rd Qu.:15.20
## Max. :1.4967 Max. :1.0380 Max. :83.50 Max. :20.20
## GNI Mat.Mor Ado.Birth Parli.F
## Min. : 2.00 Min. : 1.0 Min. : 0.60 Min. : 0.00
## 1st Qu.: 53.50 1st Qu.: 11.5 1st Qu.: 12.65 1st Qu.:12.40
## Median : 99.00 Median : 49.0 Median : 33.60 Median :19.30
## Mean : 98.73 Mean : 149.1 Mean : 47.16 Mean :20.91
## 3rd Qu.:143.50 3rd Qu.: 190.0 3rd Qu.: 71.95 3rd Qu.:27.95
## Max. :194.00 Max. :1100.0 Max. :204.80 Max. :57.50Studying the output of the summaryfunction we can see the distributions of the variables. For example, the average life expectancy is 71.65 years with a minimum of 49 and maximum of 83.5 years.
Next, we will study a graphical overview of the data. I will do this by drawing a ggpairs plot. We will also draw a correlation plot.
# visualize the 'human' variables
ggpairs(human)
# compute the correlation matrix and visualize it with corrplot
cor(human) %>% corrplot
Let’s look at the information for the distribution of each variable. All the variables are unimodally distributed (they have one peak). - Edu2.FM left-skewed distribution - Labo.FM left-skewed distribution - Life.Exp left-skewed distribution - Edu.Exp slightly left-skewed distribution, almost symmetric - Mat.Mor right-skewed distribution - Ado.Birth right-skewed distribution - Parli.F right-skewed distribution
The plot above shows that the highest correlation (in absolute values) is between the variables Mat.Mor and Life.Exp, with a negative correlation of 0.857. Other very high correlation pairs include: - Edu.Exp and Life.Exp - Mat.mor and Edu.Exp - Ado.Birth and Life.Exp - Ado.Birth and Edu.Exp - Ado.Birth and Mat.Mor
Particularly low correlations occur with the “percetange of female representatives in parliament” variable.
As we can see again, the variables have some rather high correlations between them. I feel that now, however, is a good time to remind about the fact that correlation does not equal causality, and that in social sciences particularly, the relationships betwen various variables are much more complex than often meets the eye.